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Heat Transfer Area by NTU Method

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Hot gases enter a finned tube heat exchanger at 300^{o}C and leave at 100^{o}C. It is used to heat water at a flow rate of 1 kg/s from 35^{o}C to 125^{o}C. The specific heat of exhaust hot gas is 1000 J/kg.K and the overall heat transfer coefficient based on the gas side is U_{h} = 100 W/m^{2}.K. Determine the required heat transfer surface area using the NTU method

Calculations:

Since flow configuration is not given, we shall take it as countercurrent flow.

Mass flow rate of hot gases (m_{g}) is obtained by energy balance:

1 x 4184 x (125 - 35) = m_{g} x 1000 x (300 - 100)

m_{g} = 1.8828 kg/s

C_{min} = 1.8828 x 1000 = 1882.8 W/^{o}C = C_{h}

C_{max} = 1 x 4184 = 4184 W/^{o}C = C_{c}

ε = C_{c}(T_{c,out} - T_{c,in}) / [C_{min}(T_{h,in} - T_{c,in})] = 4184 x (125 - 35) / [1882.8 x (300 - 35)]

= 0.755

C = C_{min}/C_{max} = 1882.8/4184 = 0.45

For the counter flow exchanger, effectiveness - NTU relationship is given by,

ε = {1 - exp[-N(1 -C)]} / {1 - Cexp[-N(1 - C)]}

We know ε and C; and we have to find N. This has to be solved by iteration.

0.755 = {1 - exp[-N(1 - 0.45)]} / {1 - 0.45 exp[-N(1 - 0.45)]}

0.755 = (1 - e^{-0.55N}) / (1 - 0.45e^{-0.55N})

starting with an initial assumption of N = 2,

(1 - e^{-0.55N}) / (1 - 0.45e^{-0.55N}) = 0.667 / 0.85 = 0.7845

by assuming N = 1.9,

(1 - e^{-0.55N}) / (1 - 0.45e^{-0.55N}) = 0.648 / 0.842 = 0.77

by assuming N = 1.8,

(1 - e^{-0.55N}) / (1 - 0.45e^{-0.55N}) = 0.628 / 0.833 = 0.754

Since the assumption of N = 1.8, almost balances the equation, we shall take N = 1.8 as the correct value.

The required gas surface area, A = NTU C_{min} / U = 1.8 x 1882.8 / 100 = 33.89 m^{2}

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Last Modified on: 04-Feb-2022

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